a) A có nghĩa <=> \(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}x-1\ne0\\\left(1-x\right)\left(x+1\right)\ne0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\end{matrix}\right.\)

b) Ta có:

A = \(\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\)
A = \(\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}\)

A = \(\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

A = \(\frac{x^2+x-x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

A = \(\frac{x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x-1\right)}\)

c) A = -1/2

<=> \(\frac{1}{2\left(x-1\right)}=-\frac{1}{2}\)

<=> 2(x - 1) = -2

<=> x - 1 = -1

<=> x = 0 (tmđk)

Vậy x = 0

a/ Đkxđ: \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Vậy phân thức được xác định khi \(\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

b/ \(A=\left[1+\frac{1}{x}+\frac{2}{x+1}\left(1+\frac{1}{x}\right)\right]:\frac{x^3+27}{2x}\)

\(=\left[1+\frac{1}{x}+\frac{2}{x+1}+\frac{2}{\left(x+1\right)x}\right]:\frac{\left(x+3\right)\left(x^2-3x+9\right)}{2x}\)

\(=\left[\frac{x\left(x+1\right)+\left(x+1\right)+2x+2}{\left(x+1\right)x}\right].\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}\)

\(=\frac{x^2+4x+3}{\left(x+1\right)x}.\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}=\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)x}.\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}\)

\(=\frac{2}{x^2-3x+9}\)

ĐKXĐ: \(x\ne\left\{-\frac{1}{2};\frac{1}{2};-1\right\}\)

\(B=\left(\frac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\frac{4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\left(\frac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)

\(=\frac{\left(2x^2+3x+1\right)}{\left(2x+1\right)\left(2x-1\right)}.\frac{\left(2x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{\left(x+1\right)\left(2x+1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)\left(x+1\right)\left(x^2-x+1\right)}=\frac{1}{x^2-x+1}\)

\(5,\)\(\frac{1}{5}x\left(10x-15\right)-2x\left(x-5\right)+7x\)

\(=2x^2-3x+-2x^2+10x-7x\)

\(=0\)

\(\Rightarrow\)Giá trị biểu thức không phụ thuộc vào biến x

\(6,\)\(F=5\left(x^2-3x\right)-x\left(3-5x\right)+18x+3\)

\(=5x^2-15x-3x-5x^2+18x+3\)

\(=3\)

Vậy giá trị biểu thức không phụ thuộc vào biến x 

 ( À có một số chỗ mình phải sửa đề mới đúng đó. Cậu coi lại giùm mình nha )

Cảm ơn bạn nhiều!